Results 1 to 2 of 2

Math Help - Sampling

  1. #1
    Junior Member
    Joined
    May 2006
    Posts
    37

    Sampling

    An unbiased dice is thrown once. The var(x)=35/12. The same dice is thrown 70 times.
    a) Find the probability that the mean score is less than 3.3.
    b)Find the probability that the total score exceeds 260.

    The final answer is not expected but please give working.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by kingkaisai2
    An unbiased dice is thrown once. The var(x)=35/12. The same dice is thrown 70 times.
    a) Find the probability that the mean score is less than 3.3.
    b)Find the probability that the total score exceeds 260.

    The final answer is not expected but please give working.
    I will assume that this is a question where we use the central limit theorem to
    justify treating the distribution of the mean score as normal with mean equal
    to the mean for a single throw of the die \mu=3.5 and variance equal to the variance
    of the result of a single throw divided by the sample size \sigma^2=(35/12)/70=1/24.

    a) Now a mean of less than 3.3 means 230 or less for the sum, so to correct
    for the continuity of the normal approximation we use a z-score for a mean
    of 230.5/70, which is the z-score for for a mean of 3.2928 so:

    <br />
z=\frac{3.2928-3.5}{1/\sqrt{25}}\approx -1.015<br />

    Looking this up in a normal table or using a normal calculator we have for
    a normal distributed RV we have:

    <br />
p(z<-1.015) \approx  0.155<br />

    b) There are at least two equivalent ways to do this, one another is to
    observe that the event: "total score exceeds 260.5" is the same as "the mean
    exceeds 260.5/70". I will use the latter, then the z-score for a mean of
    260.5/70 (again the extra 0.5s are continuity corrections) is:

    <br />
z=\frac{260.5/70-3.5}{1/\sqrt{24}}\approx 1.0848<br />

    Looking this up as before we find:

    <br />
p(z<1.0848) \approx 0.861<br />

    so:

    <br />
p(z>1.0848) = 1-p(z<1.0848) \approx 0.139<br />
.

    RonL

    Notes:
    i. A mean of less than 3.3 means the sum of the 70 throws is less than
    3.3x70=231. So as the normal distribution is continuous and the actual
    distribution of the sum and mean is discrete we use in the normal model a
    sum less than 230.5 to represent an actual sum of 230 or less.

    ii. Similarly in part b), a total of more than 260 is modelled as more than 260.5
    in the normal approximation.

    iii. Given the nature of the way problems are set, there is a significant chance
    that the expected answers to the two parts of this problem are the same.
    So please check the working to see if there is some slight modification which
    would make this so.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability Sampling and Sampling Methods
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: March 16th 2011, 04:41 AM
  2. sampling from a population
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 30th 2010, 05:39 AM
  3. Sampling
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 4th 2010, 11:34 PM
  4. Sampling
    Posted in the Statistics Forum
    Replies: 1
    Last Post: April 22nd 2010, 11:28 PM
  5. more on sampling
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: December 31st 2007, 09:07 AM

Search Tags


/mathhelpforum @mathhelpforum